[Haskell-cafe] Bug with QuickCheck 1.1 and GHC 6.8.2
Patrick Perry
patperry at stanford.edu
Thu Aug 14 06:15:18 EDT 2008
Actually, a much better solution is:
variant :: Int -> Gen a -> Gen a
variant v (Gen m) = Gen (\n r -> m n (rands r v))
where
rands r0 0 = r0
rands r0 n = let (r1,r2) = split r0
(n',s) = n `quotRem` 2
in case s of
0 -> rands r1 n'
_ -> rands r2 n'
Patrick
On Aug 14, 2008, at 2:17 AM, Johan Tibell wrote:
> On Thu, Aug 14, 2008 at 1:58 AM, Patrick Perry
> <patperry at stanford.edu> wrote:
>> variant :: Int -> Gen a -> Gen a
>> variant v (Gen m) = Gen (\n r -> m n (rands r !! v'))
>> where
>> v' = abs (v+1) `mod` 10000
>> rands r0 = r1 : rands r2 where (r1, r2) = split r0
>
> Note that if you have a uniformly distributed random value in [0, n)
> and take its value mod k you don't end up with another random
> uniformly distributed value unless n `mod` k == 0. Consider n = 3 and
> k = 2. What you can do is to throw away all random numbers larger than
> n - (n `mod` k) and just generate a new number. This will terminate
> with a high probability if n >> k.
>
> Cheers,
>
> Johan
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